University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises - Page 127: 59

Answer

The behaviors of the curves in the plotted graph show that the derivative of $f(x)=\sqrt x$ is $\frac{1}{2\sqrt x}$.
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Work Step by Step

The graphs of $y=\frac{1}{2\sqrt x}$ and $y=\frac{\sqrt{x+h}-\sqrt x}{h}$ are plotted and denoted below. We see that as $h$ gets smaller and smaller and approaches $0$, the curve of $y=\frac{\sqrt{x+h}-\sqrt x}{h}$ gets closer and in fact, becomes the same as the curve of $y=\frac{1}{2\sqrt x}$. We can thus deduce that on $[0,2]$: $$\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h}=\frac{1}{2\sqrt x}$$ According to the definition of derivative, $\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h}=f'(x)$ for $f(x)=\sqrt x$. Therefore, the behaviors of the curves in the plotted graph show that the derivative of $f(x)=\sqrt x$ is $\frac{1}{2\sqrt x}$.
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